Question

In a certain positive fraction , the denominator is greater than the numerator by 3. If 1 is subtracted from the numerator and the denominator both, the fraction reduces by 1/14. Find the fraction.​

Answer 1

Information provided with us:

• The denominator is greater than the numerator by 3.
• 1 is subtracted from the numerator
• 1 is also subtracted from the denominator
• Fraction is reduced by 1/14

What we have to calculate:

• We have to calculate and find out the fraction

How to solve these type of questions:

• These type of questions are of linear equations. First of all we have to assume a number be any variable eg, x or y. Then set-up an equation in terms of information given and solve it.

Performing Calculations:

Assuming that the positive fraction as,

• $\red{\boxed{ \sf{ \dfrac{y}{y + 3} }}}$

Here we have added 3 in denominator because in the question it is given denominator is greater than numerator by 3.

Subtracting 1 from numerator,

$: \longmapsto \: \sf{ \dfrac{y - 1}{y + 3} }$

Subtracting 1 from denominator,

$: \longmapsto \: \sf{ \dfrac{y - 1}{y + 3 - 1} }$

$: \longmapsto \: \sf{ \dfrac{y - 1}{y + 2} }$

We know certain positive fraction is,

$\boxed{\sf{ \dfrac{y }{y + 3} }}$

Fraction is reduced by 1/14,

$: \longmapsto \: \sf{ \dfrac{y }{y + 3} - \dfrac{1}{14} }$

Keeping it equal to when 1 was subtracted from both denominator and numerator,

$: \longmapsto \: \sf{ \dfrac{y }{y + 3} - \dfrac{1}{14} = \dfrac{y - 1}{y + 2} }$

★ Taking L.C.M.,

$: \longmapsto \: \sf{ \dfrac{14(y) }{y + 3} - \dfrac{1(y + 3)}{14} = \dfrac{y - 1}{y + 2}}$

★ Cross multiplying,

$: \longmapsto \: \sf{ \dfrac{14 \times y}{y + 3} - \dfrac{1 \times (y + 3)}{14} = \dfrac{y - 1}{y + 2}}$

$: \longmapsto \: \sf{ \dfrac{14 \times y - 1 \times (y + 3)}{14 \times (y + 3)}} = \dfrac{y - 1}{y + 2}$

$: \longmapsto \: \sf{ \dfrac{14y - 1 \times (y + 3)}{14 \times (y + 3)}} = \dfrac{y - 1}{y + 2}$

$: \longmapsto \: \sf{ \dfrac{14y - 1y - 3}{14 \times (y + 3)}} = \dfrac{y - 1}{y + 2}$

$: \longmapsto \: \sf{ \dfrac{13y - 3}{14 \times (y + 3)}} = \dfrac{y - 1}{y + 2}$

★ Changing the sides,

$: \longmapsto \: \sf{ }(13y - 3)(y + 2) = 14(y - 1)(y + 3)$

$: \longmapsto \: \sf{ }(13y - 3) \times (y + 2) = 14(y - 1) \times (y + 3)$

$: \longmapsto \: \sf{ 13y (y + 2) - 3(y + 2)= 14(y {}^{2} - y + 3y - 3)}$

$: \longmapsto \: \sf{ 13y {}^{2} + 26y - 3y - 6= 14(y {}^{2} - y + 3y - 3)}$

$: \longmapsto \: \sf{ 13y {}^{2} + 23y - 6= 14(y {}^{2} - y + 3y - 3)}$

$: \longmapsto \: \sf{ 13y {}^{2} + 23y - 6= 14(y {}^{2} + 2y - 3)}$

$: \longmapsto \: \sf{ 13y {}^{2} + 23y - 6= 14 \times (y {}^{2} + 2y - 3)}$

$: \longmapsto \: \sf{ 13y {}^{2} + 23y - 6= 14y {}^{2} + 28y - 42}$

★ Transposing the sides,

$: \longmapsto \: \sf{ 23y - 28y = 14y {}^{2} - 13y {}^{2} - 42 + 6}$

$: \longmapsto \: \sf{ - 5y = y {}^{2} - 36}$

★ Now,

$: \longmapsto \: \sf{ y {}^{2} + 5y - 36 = 0}$

★ Forming factors,

$: \longmapsto \: \sf{ y {}^{2} + 5y - 4y - 36 = 0}$

★ Grouping them,

$: \longmapsto \: \sf{ y ( y + 9) - 4(y + 9) = 0}$

$: \longmapsto \: \sf{ ( y + 9) (y - 4) = 0}$

★ Comparing them,

$: \implies \: \bf{y - 4 = 0}$

$: \implies \: \large{\boxed{\bf{y = 4}}}$

★ Finding out the fraction,

$: \longmapsto \: \tt{ \dfrac{y}{y+3}}$

• Putting the value of x as 4,

$: \longmapsto \: \tt{Fraction \: = \: \dfrac{4}{4+3}}$

$: \longmapsto \: \purple{\boxed{\bf{Fraction \: = \: \dfrac{4}{7}}}}$

Answer 2

Question:-

In a certain positive fraction , the denominator is greater than the numerator by 3. If 1 is subtracted from the numerator and the denominator both, the fraction reduces by 1/14. Find the fraction.

Given:-

Denominator of the fraction is greater than numerator by

To Find:-

• Fraction.

Solution:-

• Let the Numerator be x. and
• Denominator = x + 3.

$\sf \: Fraction = \frac{x}{x + 3}$

According to question,

$\sf \frac{x - 1}{(x + 3) - 1 } = \frac{x}{x + 3} - \frac{1}{14}$

$\sf \longrightarrow \frac{x - 1}{x + 2} = \frac{x}{x+3} - \frac{1}{14}$

$\sf\longrightarrow \frac{x}{x + 3} - \frac{x - 1}{x + 2} = \frac{1}{14}$

$\sf \longrightarrow \frac{x(x + 2) - (x - 1)(x + 3)}{(x + 3)(x + 2)} = \frac{1}{14}$

$\sf \longrightarrow \frac{x {}^{2} + 2x - (x {}^{2} + 3x - x - 3) }{(x + 3)(x + 2)} = \frac{1}{14}$

$\sf \longrightarrow \frac{{{ \cancel{x {}^{2}}}} +{{ \cancel{ 2x}}} - {{ \cancel{ x {}^{2} }}}- {{ \cancel{2x }}}+ 3 }{(x + 3)(x + 2)} = \frac{1}{14}$

➟ 42 = (x + 3) (x + 2)

➟ 42 = x² + 2x + 3x + 6

➟ x² + 5x – 36 = 0

➟ x² + 9x – 4x – 36 = 0

➟ x (x + 9) – 4 (x + 9) = 0

➟ (x + 9) (x – 4) = 0

x = – 9, 4

• Numerator = x = – 9, 4.
• Denominator = x + 3 = – 6, 7.

♤ Positive fraction = Both Numerator and denominator are natural number.

$\sf \: Fraction = \frac{4}{7}$

Answer:-

$\sf \red{Hence, Fraction = \frac{4}{7} }$

Hope you have satisfied. ⚘